Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #20
Grade 4 arithmeticnumber-theoryPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
First peel off one coin of each kind to satisfy the "at least one" rule. That fixes 41¢ in 4 coins and leaves 61¢ to share among 5 more coins. Tool #8 (Analyze the Units) gives the key restriction: nickels, dimes, and quarters all contribute multiples of 5, so the number of extra pennies must make the leftover a multiple of 5. Tool #3 (Eliminate Possibilities) then knocks out every penny count except one. With pennies pinned down, Tool #2 (Make a Systematic List) tries each possible extra-dime count and eliminates the ones that can't be completed by nickels and quarters. No equations needed — just divisibility and short bookkeeping.
Set aside one of each coin
Give one coin of each kind away first: those 4 coins fix 1+5+10+25=41¢, leaving 5 coins to make 102-41=61¢.
Spending the "at least one" requirement up front turns a four-variable puzzle into a smaller one with no minimum constraints.
4.MD.A.2Eliminate PossibilitiesFix the extra pennies
Nickels, dimes, and quarters are all multiples of 5, so the leftover's ones digit forces 1 extra penny (6 pennies won't fit in 5 coins).
Looking only at the ones digit (the unit of 1¢) forces the penny count without touching the other coins.
After one coin of each type is set aside, exactly one of the five remaining coins must be a penny.
▸ Why?
The five leftover coins are worth 61 cents together, and that total is the pennies' value (1 cent each) plus the nickels, dimes, and quarters, which by themselves always come to a multiple of 5 cents.
▸ Why?
A nickel, dime, and quarter are worth 5, 10, and 25 cents, each a whole number of fives, so any pile of them equals 5 times a whole number, which is a multiple of 5.
▸ Why?
The 61 cents is split with no gap and no overlap into the pennies' part and the other coins' part, so those two parts add back to the whole 61 cents.
▸ Why?
Since the other coins supply a multiple of 5 cents, the pennies must supply the rest of the 61 cents, and with only five coins available that rest can only be 1 cent.
▸ Why?
Taking the other coins' multiple-of-5 value away from 61 leaves the pennies' value, so the number of pennies equals 61 minus some multiple of 5.
▸ Why?
Multiples of 5 sit 5 apart (..., 55, 60, 65, ...), so the only one close enough to 61 to leave a penny count between 0 and 5 is 60, which leaves exactly one penny.
Update the leftover budget
Place that penny: now 4 coins remain to make 60¢, and they can only be nickels, dimes, or quarters.
With pennies done, every remaining coin is worth a multiple of 5¢ — the rest is pure trial on a small board.
4.MD.A.2Eliminate PossibilitiesList the cases by dimes
Test extra dimes d=0..4: only d=0 works, filled by 2 quarters and 2 nickels — every other d can't reach 60¢.
Five small cases, each settled by one quick check — exactly the situation a systematic list is built for.
4.OA.A.3Make A Systematic ListAdd the first dime back
Add the set-aside dime back: the final bag is 2 pennies, 3 nickels, 1 dime, 3 quarters = 9 coins totaling 102¢.
Don't forget to put back the dime that was set aside at the very start — the count of 1 comes from the original alone.
4.MD.A.2Eliminate PossibilitiesHand out one of each coin first to clear the "at least one" rule. The leftover 61¢ ends in a 1, so only pennies can supply that odd unit — and with just 5 coins left, there's room for exactly 1 extra penny. From there, a short check of dime counts 0 through 4 leaves only one working bag, with 1 dime total — answer (A).
- Set aside one of each coin
- Fix the extra pennies
- Update the leftover budget
- List the cases by dimes
- Add the first dime back
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